1) A semaphore consists of a counter, a waiting list, and wait() and signal() methods. Wait() decrements the counter and blocks if it becomes negative, while signal() increments the counter and resumes a blocked process if the counter becomes positive. 2) The dining philosophers problem is solved using semaphores to lock access to shared chopsticks, with one philosopher designated as a "weirdo" to avoid deadlock by acquiring locks in a different order. 3) The producer-consumer problem uses three semaphores - one to limit buffer size, one for empty slots, and one for locks - to coordinate producers adding to a bounded buffer

1. Semaphores
qA semaphore is an object that consists of a
counter, a waiting list of processes and two
methods (e.g., functions): signal and wait.
semaphore
method signal
counter
method wait
waiting list
1

2. Semaphore Method: wait
void wait(sem S)
{
S.count--;
if (S.count < 0) {
add the caller to the waiting list;
block();
}
}
qAfter decreasing the counter by 1, if the counter
value becomes negative, then
vadd the caller to the waiting list, and then
vblock itself. 2

3. Semaphore Method: signal
void signal(sem S)
{
S.count++;
if (S.count <= 0) {
remove a process P from the waiting list;
resume(P);
}
}
qAfter increasing the counter by 1, if the new
counter value is not positive, then
vremove a process P from the waiting list,
vresume the execution of process P, and return
3

4. Important Note: 1/4
S.count--; S.count++;
if (S.count<0) { if (S.count<=0) {
add to list; remove P;
block(); resume(P);
} }
qIf S.count < 0, abs(S.count) is the
number of waiting processes.
qThis is because processes are added to (resp.,
removed from) the waiting list only if the
counter value is < 0 (resp., <= 0).
4

5. Important Note: 2/4
S.count--; S.count++;
if (S.count<0) { if (S.count<=0) {
add to list; remove P;
block(); resume(P);
} }
qThe waiting list can be implemented with a
queue if FIFO order is desired.
qHowever, the correctness of a program should
not depend on a particular implementation of
the waiting list.
qYour program should not make any assumption
about the ordering of the waiting list. 5

6. Important Note: 3/4
S.count--; S.count++;
if (S.count<0) { if (S.count<=0) {
add to list; remove P;
block(); resume(P);
} }
qThe caller may be blocked in the call to wait().
qThe caller never blocks in the call to signal().
If S.count > 0, signal() returns and the
caller continues. Otherwise, a waiting process is
released and the caller continues. In this case, two
processes continue.
6

7. The Most Important Note: 4/4
S.count--; S.count++;
if (S.count<0) { if (S.count<=0) {
add to list; remove P;
block(); resume(P);
} }
qwait() and signal() must be executed
atomically (i.e., as one uninterruptible unit).
qOtherwise, race conditions may occur.
qHomework: use execution sequences to show
race conditions if wait() and/or signal() is
not executed atomically.
7

8. Three Typical Uses of Semaphores
qThere are three typical uses of semaphores:
vmutual exclusion:
Mutex (i.e., Mutual Exclusion) locks
vcount-down lock:
Keep in mind that semaphores have a counter.
vnotification:
Indicate an event has occurred.
8

9. Use 1: Mutual Exclusion (Lock)
initialization is important
semaphore S = 1;
int count = 0;
Process 1 Process 2
while (1) { while (1) {
// do something entry // do something
S.wait(); S.wait();
count++; critical sections count--;
S.signal(); S.signal();
// do something exit // do something
} }
qWhat if the initial value of S is zero?
qS is a binary semaphore (similar to a lock).
9

10. Use 2: Count-Down Counter
semaphore S = 3;
Process 1 Process 2
while (1) { while (1) {
// do something // do something
S.wait(); S.wait();
at most 3 processes can be here!!!
S.signal(); S.signal();
// do something // do something
} }
qAfter three processes pass through wait(), this
section is locked until a process calls signal().
10

11. Use 3: Notification
semaphore S1 = 1, S2 = 0;
process 1 process 2
while (1) { while (1) {
// do something // do something
S1.wait(); notify S2.wait();
cout << “1”; cout << “2”;
S2.signal(); notify S1.signal();
// do something // do something
} }
qProcess 1 uses S2.signal() to notify process
2, indicating “I am done. Please go ahead.”
qThe output is 1 2 1 2 1 2 ……
qWhat if both S1 and S2 are both 0’s or both 1’s?
qWhat if S1 = 0 and S2 = 1? 11

12. Lock Example: Dining
Philosophers
§ Five philosophers are in a
thinking - eating cycle.
§ When a philosopher gets
hungry, he sits down, picks
up two nearest chopsticks,
and eats.
§ A philosopher can eat only
if he has both chopsticks.
§ After eating, he puts down
both chopsticks and thinks.
§ This cycle continues.
12

13. Dining Philosopher: Ideas
qChopsticks are shared outer critical section
items (by two philosophers) left chop locked
and must be protected. Semaphore C[5] = 1;
qEach chopstick has a
C[i].wait();
semaphore with initial C[(i+1)%5].wait();
value 1.
has 2 chops and eats
qA philosopher calls
wait() before picks up a C[(i+1)%5].signal();
C[i].signal();
chopstick and calls
signal() to release it.
inner critical section
right chop locked 13

14. Dining Philosophers: Code
semaphore C[5] = 1;
philosopher i wait for my left chop
while (1) {
// thinking
C[i].wait(); wait for my right chop
C[(i+1)%5].wait();
// eating release my right chop
C[(i+1)%5].signal();
C[i].signal(); release my left chop
// finishes eating
}
Does this solution work?
14

15. Dining Philosophers: Deadlock!
§ If all five philosophers
sit down and pick up
their left chopsticks at
the same time, this
program has a circular
waiting and deadlocks.
§ An easy way to remove
this deadlock is to
introduce a weirdo who
picks up his right
chopstick first!
15

16. Dining Philosophers: A Better Idea
semaphore C[5] = 1;
philosopher i (0, 1, 2, 3) Philosopher 4: the weirdo
while (1) { while (1) {
// thinking // thinking
C[i].wait(); C[(i+1)%5].wait();
C[(i+1)%5].wait(); C[i].wait();
// eating // eating
C[(i+1)%5].signal(); C[i].signal();
C[i].signal(); C[(i+1)%5].signal();
// finishes eating; // finishes eating
} }
lock left chop lock right chop
16

17. Dining Philosophers: Questions
qThe following are some important questions for
you to work on.
vWe choose philosopher 4 to be the weirdo.
Does this choice matter?
vShow that this solution does not cause circular
waiting.
vShow that this solution will not have circular
waiting if we have more than 1 and less than 5
weirdoes.
qThese questions may appear as exam problems.
17

18. Count-Down Lock Example
qThe naïve solution to the
dining philosophers causes
circular waiting.
qIf only four philosophers are
allowed to sit down, no
deadlock can occur.
qWhy? If all four of them sit
down at the same time, the
right-most philosopher can
have both chopsticks!
qHow about fewer than four?
This is obvious. 18

19. Count-Down Lock Example
semaphore C[5]= 1;
semaphore Chair = 4;
get a chair
while (1) { this is a count-down lock
// thinking that only allows 4 to go!
Chair.wait();
C[i].wait();
C[(i+1)%5].wait();
// eating this is our old friend
C[(i+1)%5].signal();
C[i].signal();
Chair.signal();
} release my chair
19

20. The Producer/Consumer Problem
qSuppose we have a
circular buffer of n slots.
qPointers in (resp., out)
points to the first empty
(resp., filled) slot.
qProducer processes keep
adding info into the
buffer
qConsumer processes keep
bounded-buffer retrieving info from the
buffer. 20

21. Problem Analysis
qA producer deposits info into
Buf[in] and a consumer
retrieves info from Buf[out].
qin and out must be advanced.
qin is shared among producers.
qout is shared among consumers.
qIf Buf is full, producers should
buffer is implemented be blocked.
with an array Buf[ ] qIf Buf is empty, consumers
should be blocked.
21

22. qWe need a sem.
to protect the
buffer.
qA second sem.
to block
producers if the
buffer is full.
qA third sem. to
block
consumers if
the buffer is
empty.
22

23. Solution
no. of slots
semaphore NotFull=n, NotEmpty=0, Mutex=1;
producer consumer
while (1) { while (1) {
NotFull.wait(); NotEmpty.wait();
Mutex.wait(); Mutex.wait();
Buf[in] = x; x = Buf[out];
in = (in+1)%n; out = (out+1)%n;
Mutex.signal(); Mutex.signal();
NotEmpty.signal(); NotFull.signal();
} }
notifications
critical section
23

24. Question
qWhat if the producer code is modified as follows?
qAnswer: a deadlock may occur. Why?
while (1) {
Mutex.wait();
NotFull.wait();
Buf[in] = x;
order changed in = (in+1)%n;
NotEmpty.signal();
Mutex.signal();
}
24

25. The Readers/Writers Problem
qTwo groups of processes, readers and writers,
are accessing a shared resource by the following
rules:
vReaders can read simultaneously.
vOnly one writer can write at any time.
vWhen a writer is writing, no reader can read.
vIf there is any reader reading, all incoming
writers must wait. Thus, readers have higher
priority.
25

26. Problem Analysis
qWe need a semaphore to block readers if a
writer is writing.
qWhen a writer arrives, it must be able to know
if there are readers reading. So, a reader count
is required which must be protected by a lock.
qThis reader-priority version has a problem:
bounded waiting condition may be violated if
readers keep coming, causing the waiting
writers no chance to write.
26

27. qWhen a reader comes
in, it increase the
count.
qIf it is the 1st reader,
waits until no writer is
writing,
qReads data.
qDecreases the counter.
qNotifies the writer
that no reader is
reading if it is the last.
27

28. qWhen a writer
comes in, it waits
until no reader is
reading and no
writer is writing.
qThen, it writes data.
qFinally, notifies
readers and writers
that no writer is in.
28

29. Solution
semaphore Mutex = 1, WrtMutex = 1;
int RdrCount;
reader writer
while (1) { while (1) {
Mutex.wait();
RdrCount++;
if (RdrCount == 1) blocks both readers and writers
WrtMutex.wait(); WrtMutex.wait();
Mutex.signal();
// read data // write data
Mutex.wait();
RdrCount--;
if (RdrCount == 0)
WrtMutex.signal(); WrtMutex.signal();
Mutex.signal();
} } 29